Question #263208

A battery of 10V emf and internal resistance of 0.50Ω is connected in parallel with another battery of 12V and internal resistance of 20Ω. What is the load current?

1
Expert's answer
2021-11-12T08:13:04-0500

The load current is


I=V1−V2r1+r2=12−1020+0.5=0.098 A.I=\frac{V_1-V_2}{r_1+r_2}=\frac{12-10}{20+0.5}=0.098\text{ A}.


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